Approx. 112 mL.
1.68 drops
Thevolume is 24,12 mL.
36.5 g ZnS
2Mg + O2 ==> 2MgO Balanced Equation4.03 g Mg x 1 mole Mg/24.3 g = 0.166 moles Mg present in 4.03 g Mg.moles O2 required = 0.166 moles Mg x 1 mole O2/2 moles Mg = 0.083 moles O2 needed.At STP 1 mole occupies 22.4 L, thusVolume of O2 required = 0.083 moles x 22.4 L/mole = 1.56 L x1000 ml/L = 1859 mlsSince 4.03 g has only 3 significant figures, the correct answer should be 1860 milliliters.
The volume of N2O5 obtained is 585 mL.
The reaction equation is: 2H2 + O2 = 2H2O So one need two volume part of hydrogen and one volume part of oxygen to form water.
The answer is 152 g oxygen.
g
65
Three atoms of oxygen are required to react with each two atoms of aluminum to form the most common product of reaction between oxygen and aluminum. Therefore, 0.75 mole of oxygen atoms will be required to react with 0.5 mole of aluminum atoms. The atomic weight of oxygen is 15.999; therefore, the mass will be (0.75)(15.999) = 12 grams of oxygen, to the maximum possibly justified number of significant digits.
7.20L
5g of oxygen are required to completely react with 10g of hydrogen resulting in water with no net losses. Or exactly 2:1 = H20 (2 hydrogen < > 1 oxygen)
1.68 drops
Thevolume is 24,12 mL.
Approximately twice as much volume of hydrogen as of oxygen: Both gases are diatomic and nearly ideal at normal temperature and pressure, and the atomic ratio of hydrogen to oxygen in water is 2.
36.5 g ZnS
3